zoukankan      html  css  js  c++  java
  • B. Amr and The Large Array(Codeforces Round #312 (Div. 2)+找出现次数最多且区间最小)

    B. Amr and The Large Array
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Amr has got a large array of size n. Amr doesn't like large arrays so he intends to make it smaller.

    Amr doesn't care about anything in the array except the beauty of it. The beauty of the array is defined to be the maximum number of times that some number occurs in this array. He wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.

    Help Amr by choosing the smallest subsegment possible.

    Input

    The first line contains one number n (1 ≤ n ≤ 105), the size of the array.

    The second line contains n integers ai (1 ≤ ai ≤ 106), representing elements of the array.

    Output

    Output two integers l, r (1 ≤ l ≤ r ≤ n), the beginning and the end of the subsegment chosen respectively.

    If there are several possible answers you may output any of them.

    Sample test(s)
    input
    5
    1 1 2 2 1
    
    output
    1 5
    input
    5
    1 2 2 3 1
    
    output
    2 3
    input
    6
    1 2 2 1 1 2
    
    output
    1 5
    Note

    A subsegment B of an array A from l to r is an array of size r - l + 1 where Bi = Al + i - 1 for all 1 ≤ i ≤ r - l + 1

       题意:给出一个序列,然后找出出现次数最多,但区间占用长度最短的区间左右值。

    点击打开链接

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define N 1000001
    
    using namespace std;
    
    int n,m;
    
    struct node
    {
        int x;
        int y;
        int ans;
        int cnt;
    }q[1000100];
    
    bool cmp(node a,node b)
    {
        if(a.cnt == b.cnt)
        {
            return a.ans < b.ans;
        }
        return a.cnt > b.cnt;
    }
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            int mm;
            for(int i=0;i<=N;i++)
            {
                q[i].cnt = 0;
            }
            int maxx = 0;
            int a;
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a);
                if(mm<a)
                {
                    mm = a;
                }
                if(q[a].cnt == 0)
                {
                    q[a].x = i;
                    q[a].y = i;
                    q[a].ans = 0;
                    q[a].cnt++;
                }
                else
                {
                    q[a].cnt++;
                    q[a].y = i;
                    q[a].ans = q[a].y - q[a].x;
                }
                if(maxx<q[a].cnt)
                {
                    maxx = q[a].cnt;
                }
            }
            sort(q,q+N,cmp);
            printf("%d %d
    ",q[0].x,q[0].y);
    
        }
        return 0;
    }


  • 相关阅读:
    Docker03-镜像
    Docker02:Centos7.6安装Docker
    Docker01-重要概念
    WEB开发新人指南
    Lpad()和Rpad()函数
    Unable to find the requested .Net Framework Data Provider. It may not be installed
    redis自动过期
    redis简单的读写
    redis的安装
    Ajax缓存,减少后台服务器压力
  • 原文地址:https://www.cnblogs.com/tlnshuju/p/7094851.html
Copyright © 2011-2022 走看看